(II) Assume the supports of the uniform cantilever shown in Fig. 9-76 (m = 2900 kg) are made of

step 1 So once again, I'll welcome to a new problem. We have a candidever running across towards the li

(II) Assume the supports of the uniform cantilever shown in...

Key Concepts

Cantilever Beam Analysis

A cantilever beam is a structural element that is fixed at one end and free at the other, creating a non-symmetrical load distribution with maximum bending moments and shear forces at the fixed end. Understanding how these forces develop along the length of the beam is essential in determining the stresses that will act on the supports or the beam itself.

Load Conversion and Static Equilibrium

In structural design, it is necessary to convert masses into forces using gravitational acceleration, and then apply the principles of static equilibrium to ensure that the sum of forces and moments in the system balance. This approach forms the basis for evaluating the loads that the material must safely resist.

Stress Analysis in Structural Members

Stress analysis involves determining the internal forces (such as bending stresses, axial stresses, and shear stresses) within a material or a structural component due to the applied external loads. The calculated stress is then compared with the material’s allowable stress to ensure that the design remains safe under working conditions.

Factor of Safety

The factor of safety is a design parameter used to provide a margin between the expected maximum load and the load at which failure might occur. By multiplying estimated stresses or reducing allowable loads by the safety factor, engineers ensure that the structure remains reliable even under unforeseen increases in loading or material imperfections.

Material Properties and Design of Wood

Wood, as a construction material, has specific mechanical properties, including compressive strength, tensile strength, and modulus of elasticity. In the design process, these properties must be carefully considered to ensure that the wooden supports can handle the converted loads multiplied by the safety factor while remaining within the safe stress limits prescribed by material standards.

Transcript

00:01 So once again, i'll welcome to a new problem.

00:04 We have a candidever running across towards the light, and it has two supports.

00:12 There is one support that's on the right edge, and then there's another support that's close to the center, but not exactly at the center.

00:24 I think i want to place it that way.

00:27 The distance between these two supports happens to be 20 meters, so from right there.

00:34 Up until there you have 20 meters.

00:39 So there are two forces.

00:40 The right support has a reaction are a because we're calling it a and then the that's the left support sorry and then the right support as a second reaction we're calling it r b.

00:55 This is v.

00:56 The distance from the middle support or the right support up until the edge right there that's early meters that's the distance the weight of the carnivir adds through the middle the weight of the carnivir adds to the middle well the mass itself m is 2 ,900 kilograms that's the mass and then we can always get the weight of this so mg will be 2 ,900 900, 2 ,900.

01:51 Oh, you know what? i want to have it in kilograms.

01:54 So 2 ,900 kilograms times 9 .81 meters per second square.

02:01 That's going to be the weight like mg of the whole thing.

02:04 I don't want to think about the weight yet, but i just know that it's mg at the moment.

02:10 The distance from the center up until the right edge is 25 meters.

02:16 And then from this center, up until all the way up until this side is also 25 meters and that's from the center itself our goal that's what's given in the problem our goal is to find the cross -sectional area so you know for the left one of the a the cross -sectional area will be a here that's the cross -sectional area of a and then also you want to think about the cross -sectional area of b which is this one so if you cut across b what's going to be the cross -sectional area, the system is in equilibrium.

03:06 The system is in equilibrium.

03:13 So that simply means that the sum of forces in the wide direction is 0.

03:24 So ra plus rb, because they're pointing upwards, so they're both positive, and then minus mg is 0.

03:35 So this is mg is pointing downwards, so it's negative.

03:40 And so ra plus rb equals to mg.

03:48 So the reaction at point a is ra and the reaction at point b is rb.

03:58 The sum of talks around a is zero.

04:05 And so if you think about b itself, b is going counterclockwise relative to the point and then the weight is going clockwise relative to the point a so the weight is going to have a negative torque and the reaction at b will have a positive torque but then the distance from the reaction at b up until the left edge a is 20 meters so we have r b times 20 meters r b times 20 meters it's positive because it has a positive torque going counterclockwise then minus m g times 25 meters uh that's equivalent to zero now this is going to help us solve for r b so r b equals to md times 25 meters mg is the weight then we have to divide both sides by 20 meters and that gives us r b in terms of the weight so 1 .25 times the weight.

05:29 That's the fraction of the weight that rb is kind of like supporting.

05:34 Then we're going to go to the next page.

05:37 Since we know that the two reactions have to be equivalent to the weight, otherwise the system is going to collapse.

05:46 You can solve for the left reaction, which is mg minus rb, using algebra.

05:55 But then since from this previous page, you see rb is 1 .25.

06:00 Of the weight.

06:02 So we can always do those replacements and have this is 1 .25 times the weight.

06:09 And so our a is negative 0 .25 of the weight.

06:16 The negative just means that it's in the opposite direction as the weight.

06:20 So don't think about that too much.

06:24 R .b or r .a now is 0 .25 of the weight.

06:30 And the second reaction is 1 .25.

06:33 Of the weight.

06:36 Those are the two reactions.

06:40 A member will grow from the beginning was to get these two cross -sectional areas.

06:45 So we're using equilibrium as a way of helping us off for the two reactions.

06:51 But we don't stop there because we have to think about the stress at a.

06:55 The reason why we think about the stress at a is because it's going to give us a relationship that has the area.

07:06 That's what we're looking for.

07:08 We're looking for the cross -sectional area.

07:10 So if you can figure out a formula, stress formula, that has an area, it's going to help us solve the problems.

07:19 So the reaction at a is 0 .25 of the weight, and then that's going to be divided by the area at a.

07:30 There's another concept that will be helpful in solving this problem, and this is what we call the safety factor.

07:42 And the safety factor is denoted by n.

07:48 And in this case, for part a, the safety factor n is the same as 40 times 10 to the 6 newton per meter squared in this case.

08:04 And it's the relationship between the ultimate stress...

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